How many total subsets can be made of the set: {A,L,V,I,N}?

nckoliss

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Feb 5, 2009
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i am on our UIL team for math and i came across a question in our competition that not evan my math teacher knows how to do. can anyone help me. the question geos like this-

How many total subsets can be made of the set: {A,L,V,I,N}.

the competition is over and i won't get to do it again till next year, but i still would like to know how to do. please help
here is another one i had that i need help on too.

the number of elements in the power set of {B,E,L,T,O,N}.

all i realy want to know is how to do it and can you tell me what it's called.
 
Re: help please i don't evan know what it's called

The power set is the collection of all subsets of a given set, including the set itself and the null set. A set with \(\displaystyle N\) elements gives a power set of \(\displaystyle 2^N\) members.
 
It means you can make \(\displaystyle 2^{n}\) subsets from a set with n elements. Yours has 5 elements, so \(\displaystyle 2^{5}=32\) subsets can be made.
 
so if i understand this right, each letter is an element, and that number becomes an exponent over 2, and that is my answer?
 
In case you are wondering why this is so:
A set with n number of elements can have numerous subsets (a subset is all possible combination you can think of made with the elements of the given set):
1. A set with no element (also called the NULL set) {}
2. Sets with 1 element like {A}, {L}, {V} and so on in your case
3. Sets with 2 elements like {A,L}, {A,V}, {A,I}, {L,V} and so on in your case
4. Sets with 3 elements like {A,L,V}, {L,V,N}, {A,I,N} and so on in your case
.
.
.
.
n. A single set with all the elements of the given set {A,L,V,I,N}

So the total number of subsets = [sup:eek:qr0ggo6]n[/sup:eek:qr0ggo6]C[sub:eek:qr0ggo6]0[/sub:eek:qr0ggo6]+[sup:eek:qr0ggo6]n[/sup:eek:qr0ggo6]C[sub:eek:qr0ggo6]1[/sub:eek:qr0ggo6]+[sup:eek:qr0ggo6]n[/sup:eek:qr0ggo6]C[sub:eek:qr0ggo6]2[/sub:eek:qr0ggo6]+......+[sup:eek:qr0ggo6]n[/sup:eek:qr0ggo6]C[sub:eek:qr0ggo6]n[/sub:eek:qr0ggo6]
As per the Binomial Theorem (I guess that's what it is called) the above sum culminates to 2[sup:eek:qr0ggo6]n[/sup:eek:qr0ggo6]
 
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